线性代数: 第2题的(1)(2)(3)小题,需要详细过程,急,求帮忙
1个回答
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|λI-A| =
λ+1 -2 -2
-2 λ+1 2
-2 2 λ+1
=
λ+1 -2 -2
0 λ-1 -(λ-1)
-2 2 λ+1
=(λ-1)
λ+1 -2 -2
0 1 -1
-2 2 λ+1
=(λ-1)
λ+1 0 -2
0 2 -1
-2 1-λ λ+1
=(λ-1){(λ+1)[2(λ+1)+(1-λ)]-8}
= (λ-1)[(λ+1)(λ+3)-8] =(λ-1)(λ+5)(λ-1) = 0
解得λ=1(两重),-5
(2)
(A^-1)*=|A^-1|(A^-1)^-1
=A/|A|
而|A|=1×1×(-5)=-5
特征值是
1/(-5),1/(-5),-5/(-5)
即
-1/5,-1/5,1
(3)
E+A^-1特征值是
1+1/(-1)=0
1+1/(-1)=0
1+1/(-5)=4/5
λ+1 -2 -2
-2 λ+1 2
-2 2 λ+1
=
λ+1 -2 -2
0 λ-1 -(λ-1)
-2 2 λ+1
=(λ-1)
λ+1 -2 -2
0 1 -1
-2 2 λ+1
=(λ-1)
λ+1 0 -2
0 2 -1
-2 1-λ λ+1
=(λ-1){(λ+1)[2(λ+1)+(1-λ)]-8}
= (λ-1)[(λ+1)(λ+3)-8] =(λ-1)(λ+5)(λ-1) = 0
解得λ=1(两重),-5
(2)
(A^-1)*=|A^-1|(A^-1)^-1
=A/|A|
而|A|=1×1×(-5)=-5
特征值是
1/(-5),1/(-5),-5/(-5)
即
-1/5,-1/5,1
(3)
E+A^-1特征值是
1+1/(-1)=0
1+1/(-1)=0
1+1/(-5)=4/5
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